One end of a string of length `L` is tied to the ceiling of a lift acceleration upwards with an acceleration `2g`. The other end of the string is free. The linear mass density of the string varies linearly from `0 to lambda` from bottom to top. Choose the correct option(s)

Options

A. The velocity of the wave in the string will be 0.
B. The acceleration (relative to string) of the wave on the string will be `3g//4` every where.
C. The time taken by a pulse to reach from bottom to will be `sqrt(8L//3g)`.
D. The time taken by a pulse to reach from bottom to top will be `sqrt(4L//3g)`.

Explanation

To solve the problem regarding the string tied to a ceiling in an accelerating lift, we need to analyze the forces acting on the string due to its variable mass density and the acceleration of the lift.

### Step-by-Step Solution:

1. **Understanding the System**:
- The string has a length \( L \) and a linear mass density \( \lambda \) that varies from \( 0 \) at the bottom to \( \lambda \) at the top.
- The lift is accelerating upwards with an acceleration \( 2g \), where \( g \) is the acceleration due to gravity.

2. **Determine the Mass of the String**:
- The mass density varies linearly, so we can express the mass density as:
\[ \mu(y) = \frac{\lambda}{L} y \]
where \( y \) is the distance from the bottom of the string.
- The total mass \( m \) of the string can be calculated by integrating the mass density along the length of the string:
\[ m = \int_0^L \mu(y) \, dy = \int_0^L \frac{\lambda}{L} y \, dy = \frac{\lambda}{L} \cdot \frac{L^2}{2} = \frac{\lambda L}{2} \]

3. **Calculate the Tension in the String**:
- The tension \( T \) at a distance \( y \) from the bottom of the string can be expressed as:
\[ T(y) = m(y) \cdot (2g) = \left(\frac{\lambda}{L} y\right) \cdot (2g) = \frac{2\lambda g}{L} y \]
- Here, \( m(y) \) is the mass of the string below point \( y \).

4. **Determine the Total Force**:
- The total force acting on the string due to the lift's acceleration is:
\[ F = T(y) - m(y)g = \frac{2\lambda g}{L} y - \left(\frac{\lambda}{L} y\right) g = \left(\frac{2\lambda g}{L} y - \frac{\lambda g}{L} y\right) = \frac{\lambda g}{L} y \]
- This indicates that the tension increases linearly with \( y \).

5. **Conclusion**:
- The tension in the string varies linearly from \( 0 \) at the bottom to \( \frac{2\lambda g}{L} L = 2\lambda g \) at the top.
- The correct options would depend on the specific choices given in the question, but we have established the behavior of the tension in the string under the given conditions.